Algorithm Design

24
Chapter 1 Introduction: Some Representative Problems
The interest, naturally, was in finding a way of assigning each student
to at most one hospital, in such a way that all available positions in all
hospitals were filled. (Since we are assuming a surplus of students, there
would be some students who do not get assigned to any hospital.)
We say that an assignment of students to hospitals is stable ff neither
of the following situations arises.
¯ First type of instability: There are students s and s’, and a hospital h,
so that
- s is assigned to h, and
- s’ is assigned to no hospital, and
- h prefers s’ to s.
° Second type of instability: There are students s and s ~, and hospitals
t~ and h’, so that
- s is assigned to h, and
s’ is assigned to tff, and
- t~ prefers s’ to s, and
- s’ prefers tt to h’.
So we basically have the Stable Matching Problem, except that (i)
hospitals generally want more than one resident, and (ii) there is a surplus
of medical students.
Show that there is always a stable assignment of students to hospitals,
and give an algorithm to find one.
The Stable Matching Problem, as discussed in the text, assumes that all
men and women have a fully ordered list of preferences. In this problem
we will consider a version of the problem in which men and women can be
indifferent between certain options. As before we have a set M of n men
and a set W of n women. Assume each man and each woman ranks the
members of the opposite gender, but now we allow ties in the ranking.
For example (with n = 4), a woman could say that ml is ranked in first
place; second place is a tie between mz and m3 (she has no preference
between them); and m4 is in last place. We will say that tv prefers m to m’
if m is ranked higher than m’ on her preference list (they are not tied).
With indifferences in the ranldngs, there could be two natural notions
for stability. And for each, we can ask about the existence of stable
matchings, as follows.
(a) A strong instability in a perfect matching S consists of a man m and
a woman tv, such that each of m and tv prefers the other to their
partner in S. Does there always exist a perfect matching with no
(b)
Exercises
strong instability? Either give an example of a set of men and women
with preference lists for which every perfect matching has a strong
instability; or give an algorithm that is guaranteed to find a perfect
matching with no strong instability.
A weak instability in a perfect matching S consists of a man m and
a woman tv, such that their partners in S are tv’ and m’, respectively,
and one of the following holds:
- m prefers u~ to ui, and tv either prefers m to m’ or is indifferent
be~veen these two choices; or
u~ prefers m to m’, and m either prefers u~ to u3’ or is indifferent
between these two choices.
In other words, the pairing between m and tv is either preferred
by both, or preferred by one while the other is indifferent. Does
there always exist a perfect matching with no weak instability? Either
give an example of a set of men and women with preference lists
for which every perfect matching has a weak instability; or give an
algorithm that is guaranteed to find a perfect matching with no weak
instability.
6. Peripatetic Shipping Lines, inc., is a shipping company that owns n ships
and provides service to n ports. Each of its ships has a schedule that says,
for each day of the month, which of the ports it’s currently visiting, or
whether it’s out at sea. (You can assume the "month" here has m days,
for some m > n.) Each ship visits each port for exactly one day during the
month. For safety reasons, PSL Inc. has the following strict requirement:
(t) No two ships can be in the same port on the same day.
The company wants to perform maintenance on all the ships this
month, via the following scheme. They want to truncate each ship’s
schedule: for each ship Sg, there will be some day when it arrives in its
scheduled port and simply remains there for the rest of the month (for
maintenance). This means that S~ will not visit the remaining ports on
its schedule (if any) that month, but this is okay. So the truncation of
S~’s schedule will simply consist of its original schedule up to a certain
specified day on which it is in a port P; the remainder of the truncated
schedule simply has it remain in port P.
Now the company’s question to you is the following: Given the schedule
for each ship, find a truncation of each so that condition (t) continues
to hold: no two ships are ever in the same port on the same day.
Show that such a set of truncations can always be found, and give an
algorithm to find them.
25